Jul 6 The Cigar Puzzle

Once again with have a Dudeney puzzle. This one I picked particularly for its Victorian appeal of having gentlemen's clubs and partially it is an important idea in game theory as explored below the answer. But first the puzzle in its original words:

“I once propounded the following puzzle in a London club, and for a considerable period it absorbed the attention of the members. They could make nothing of it, and considered it quite impossible of solution. And yet, as I shall show, the answer is remarkably simple.

Two men are seated at a square-topped table. Once places an ordinary cigar (flat at one end, pointed at the other) on the table, then the other does the same, and so on alternately, a condition being that no cigar shall touch another. Which player should succeed in placing the last cigar? The size of the table top and the size of the cigar are not given, but in order to exclude the ridiculous answer that the table might be so diminutive as only to take one cigar, we will say that the table must not be less that 2 feet square and the cigar must not be more that 4.5 inches long. With those restrictions you may take any dimensions you like. Of course we assume that all the cigars are exactly alike in every respect. Should the first player, or the second player, win?”

Two hints and an answer below.

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Hint 1: the shape of the cigars is important. How can you abuse them?

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Hint 2: we need a general strategy which will work on all squares. What is the important first move?

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Answer in Dudeney’s words:

“Not a single member of the club mastered this puzzle, and yet I shall show that it is so simple that the merest child can understand its solution- when it is pointed out to him! [Once again I cry sexism.] The large majority of my friends expressed their entire bewilderment. Many considered that ‘the theoretical result, in any case, is determined by the relationship between the table and the cigars;' others, regarding it as a problem in the theory of Probabilities, arrived at the conclusion that the chances are slightly in favour of the first or second player, as the case may be. One man took a table and a cigar of particular dimensions, divided the table into equal sections, and proceeded to make the two players fill up these sections so that the second player should win. But why should the first player be so accommodating? At any stage he has only to throw down a cigar obliquely across several of these sections entirely to upset Mr. 2's calculations! We have to assume that each player plays the best possible; not that one accommodates the other.

The theories of some other friends would be quite sound if the shape of the cigar were that of a torpedo- perfectly symmetrical and pointed at both ends

I will show that the first player should infallibly win, if he always plays in the best possible manner. Examine carefully the following diagram, No. 1, and all will be clear.

The first player must place his cigar on end in the exact centre of the table, as indicated by the little circle. Now whatever the second player should do throughout, the first player must always repeat it in an exactly diametrically opposite position. Thus, if the second player places a cgar at A, I put one at AA; he places one at B, I put one at BB; and so on until no more cigars cane be placed without touching. As the cigars are supposed to be exactly alike in every respect, it is perfectly clear that for every move that the second player may choose to make, it is possible exactly to repeat it on the line drawn through the centre of the table.

The second diagram will serve to show why the first cigar must be placed on end. ( And here I will say that the first cigar that I selected from the box I was able to stand on end, and I am allowed to assume that all over cigars would do the same.) If the first cigar were were placed on its side, as at F, then the second player could place a cigar as at G- as near as possible, but not actually touching F. Now, in this position you can cannot repeat his play on the opposite side in the same relation to the centre, intersects, or lies on the top of, F, whereas the cigars are not allowed to touch. You must therefore put the cigar farther away from the centre, which would result in your having insufficient room between the centre and the bottom left-hand corner to repeat everything that the other player would do between G and the top right-hand corner. therefore the result would not be a certain win for the first player."

Back to Alaric speaking. This breaking of symmetry is important to a lot of games. When people start playing chess at school there will we a period where copying the other person's moves seems like a strategy which may be of use to get a draw against a stronger player. The trick to breaking the pattern is to play moves which can't be reciprocated. To give an example which can easily be imagined, if you were to move a queen into one of the central spaces of the board, then your opponent does the same. You can simply use your queen to take their's and now they have no queen to try the same thing. There are actually very long strings of moves in some chess openings where the book move is to keep things symmetrical. The English opening has loads of them. The trick is to know when to break the symmetry to your advantage. In case of the cigar problem the trick was to force the symmetry to your advantage.

Just a final word on the relevance to Go. Go boards are always an oddxodd square. A strategy attempted by a lot of beginners is similar to the cigar problem; they place their first stone in the middle of the board and then copy the other person's move. Why doesn't that work with Go? (If you don't know the rules of Go, then maybe it is time to learn.)